[tex]\huge\bold{Given:}[/tex]
Length of the base = 8
Length of the hypotenuse = 17
[tex]\huge\bold{To\:find:}[/tex]
The length of the third side ''[tex]x[/tex]".
[tex]\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}[/tex]
[tex]\longrightarrow{\purple{x\:=\: 15}}[/tex]
[tex]\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\red{:}}}}}[/tex]
Using Pythagoras theorem, we have
(Perpendicular)² + (Base)² = (Hypotenuse)²
[tex]\longrightarrow{\blue{}}[/tex] [tex]{x}^{2}[/tex] + (8)² = (17)²
[tex]\longrightarrow{\blue{}}[/tex] [tex]{x}^{2}[/tex] + 64 = 289
[tex]\longrightarrow{\blue{}}[/tex] [tex]{x}^{2}[/tex] = 289 - 64
[tex]\longrightarrow{\blue{}}[/tex] [tex]{x}^{2}[/tex] = 225
[tex]\longrightarrow{\blue{}}[/tex] [tex]x[/tex] = [tex]\sqrt{225}[/tex]
[tex]\longrightarrow{\blue{}}[/tex] [tex]x[/tex] = [tex]15[/tex]
Therefore, the length of the missing side [tex]x[/tex] is [tex]15[/tex].
[tex]\huge\bold{To\:verify :}[/tex]
[tex]\longrightarrow{\green{}}[/tex] (15)² + (8)² = (17)²
[tex]\longrightarrow{\green{}}[/tex] 225 + 64 = 289
[tex]\longrightarrow{\green{}}[/tex] 289 = 289
[tex]\longrightarrow{\green{}}[/tex] L.H.S. = R. H. S.
Hence verified.
[tex]\huge{\textbf{\textsf{{\orange{My}}{\blue{st}}{\pink{iq}}{\purple{ue}}{\red{35}}{\green{♨}}}}}[/tex]
